Letters are arranged in boxes

Question

Ways =(selecting one box from upper and lower rows)(selecting four boxes out of remaining)(number of ways arranging six letters in these boxes)

$$={3\choose 1}{3\choose 1}{6 \choose 4}\cdot6!$$

Answer 1

[\begin{align} & There\text{ are 3 cases to be considered : }\left( 1,1,1,3 \right)\text{ ,}\left( 2,1,1,2 \right)\text{ }and\text{ }\left( 3,1,1,1 \right) \\ & \underline{case\text{ 1}}\text{ : }top\text{ }row\text{ }1:6C1*3 \\ & \text{ }second\text{ }row\text{ }1:5C1 \\ & ~\text{ }third\text{ }row\text{ }1:\text{ }4C1 \\ & ~\text{ }fourth\text{ }row\text{ }3\text{ : }3C3*3! \\ & Soln~\text{ : }6C1*3*5C1*4C1*\text{ }3C3*3!=2160 \\ & \text{-------------------------------------------------------------------------------------} \\ & \underline{case\text{ 2}}\text{ : }top\text{ }row\text{ }1:6C2*3P2 \\ & \text{ }second\text{ }row\text{ }1:4C1 \\ & ~\text{ }third\text{ }row\text{ }1:\text{ }3C1 \\ & ~\text{ }fourth\text{ }row\text{ }3\text{ : }3C3*3P2 \\ & Soln~\text{ : }6C2*3P2*4C1*3C1*3C3*3P2=6480 \\ & ------------------------------------------ \\ & \underline{case\text{ 3}}\text{ : }top\text{ }row\text{ }1:6C3*3! \\ & \text{ }second\text{ }row\text{ }1:3C1 \\ & ~\text{ }third\text{ }row\text{ }1:\text{ }2C1 \\ & ~\text{ }fourth\text{ }row\text{ }3\text{ : }1C1*3 \\ & Soln~\text{ : }6C3*3!*3C1*2C1*\text{ }1C1*3=2160 \\ & ------------------------ \\ & Total\text{ }Soln=case1+\text{ }case2+case3=2160+4680+2160=10800 \\ \end{align}]

Answer 2

No, that is both over counting and does not ensure that all rows have at least one box selected.

You need to select, the boxes in the 2 middle rows and either:

• 1 from 3 in the top row and all 3 in the bottom row, xor
• all 3 in the top row and 1 from 3 in the bottom row, xor
• 2 from 3 in the top row and 2 from 3 in the bottom row

Then when you have selected the boxes, arrange the 6 letters within them.

Count the ways.

$10800$